Is $x^2-y$ squarefree?

Question

Let $x, y\in\mathbb{Z}$, $x^2>y$ and $x \equiv2$ mod $4$. $ $ Is it true that $x^2-y$ is squarefree? How can I prove or disprove this?

Answer 1

Use some small example, like $x=6=4+2$, $y=27$. Then $x^2=36>27=y$, and $x^2-y=9=3^2$, which is a square.

Answer 2

$y$ can be anything less than $x^2$?

Then let $y = x^2 - $ something not square-free.

Example: Let $x = 6\equiv 2 \mod 4$. $x^2 = 36$. The numbers $4,8,12,16,20,24,28,32,9,18,27,25$ are all not square-free. Let $y$ be $36$ minus any of those. So $y = 36-12 = 24$.

Then $x^2 - y= 36-24 = 12 = 3*4$ is not square free.